Let $x^2-y^2=28$. What is the value of $\dfrac{d^2y}{dx^2}$ at the point $(8,-6)$ ? Give an exact number.
Explanation: Notice that the equation defines $y$ implicitly—we don't have an explicit expression for $y$ in terms of $x$. So we will have to use implicit differentiation. If we differentiate the equation once, we will be able to get an expression for $\dfrac{dy}{dx}$. Then we can differentiate the equation again to get an expression for $\dfrac{d^2y}{dx^2}$. Let's start by finding $\dfrac{dy}{dx}$. $\dfrac{dy}{dx}=\dfrac{x}{y}$ Now we can differentiate $\dfrac{dy}{dx}$ to find $\dfrac{d^2y}{dx^2}$. $\dfrac{d^2y}{dx^2}=\dfrac{y^2-x^2}{y^3}$ Finally, let's plug ${x=8}$ and ${y=-6}$ into the expression we got: $\begin{aligned} \left.\dfrac{ y^2- x^2}{ y^3}\right\rvert_{({8},{-6})}&=\dfrac{({-6})^2-({8})^2}{({-6})^3} \\\\ &=\dfrac{7}{54} \end{aligned}$ In conclusion, the value of $\dfrac{d^2y}{dx^2}$ at the point $(8,-6)$ is $\dfrac{7}{54}$.